Osculator for 13 is 4 (See note). But this time our osculator is
not negative (as in case of 7). It is 'one-more' osculator. So, the working
will be different now. This can be seen in the following examples.
Ex 1 : Is 143 divisible by 13 ?
Soln : 143 : 14 + 3 x 4 = 26
Since 26 is divisible by 13, the number 143 is also divisible by 13.
or,
This working may further be simplified as
Step I : 1 4 3
16 [4 x 3 (from 143) + 4 (from 143)]
Step II: 1 4 3
26/16 [4 x 6 (from 16) +1 ( from 16) +1 ( from 143) = 26]
As 26 is divisible by 13, the number is also divisible by 13.
Note : The working of second method is also very systematic. At the
same time it is more acceptable because it has less writing work.

Ex 3: Check the divisibility of 6944808 by 13.
Soln : 6 9 4 4 8 0 8
39/18/12/41/19/32
4x8+0 = 32
4x2+3+8 = 19
4x9+1+4 = 41
4x1+4+4 = 12
4x2+1+9 = 18
4x8+1+6 = 39
Since 39 is divisible by 13, the given number is divisible by 13.Note : (1) This method is applicable for "one-more" osculator only. So
we can't use this method in the case of 7.
(2) This is a one-line method and you don't need to write the calculations
during exams. These are given merely to make you understand well.